# Local isometry implies covering map: nonempty boundary case [closed]

The following theorem is well known in the literature:

Let $$M$$ and $$N$$ be riemannian manifolds and let $$f : M \to N$$ be a local isometry. If $$M$$ is complete and $$N$$ is connected, then $$f$$ is a covering map.

My question is: does the same theorem hold when we assume that $$M$$ and $$N$$ are now riemannian manifolds with boundary?

• Just consider the inclusion $[0,1]\subset [0,2]$. – YCor Dec 10 '19 at 13:14

I think it does not hold: For example, let $$M$$ be any complete Riemannian manifold with connected, totally geodesic boundary, and let $$N$$ be its double—the disjoint union of two copies of $$M$$ glued along their boundary. The inclusion map $$M\to N$$ is a local isometry, but is not a covering map.
• But in this case $N$ does not have a boundary, which is assumed in my question. – Eduardo Longa Dec 10 '19 at 2:15
• @EduardoLonga: could one not use the same idea for a riemannian manifold with two isomorphic boundary components, gluing just one? More specifically, I'm thinking of a cylinder $[0, 1] \times \mathbb{S}^1$, and gluing two of these: one along $\{0\} \times \mathbb{S}^1$ and the other along $\{1\} \times \mathbb{S}^1$. – Eric Canton Dec 10 '19 at 2:18
• Or, less complicated: take $f: M \to N$ to be the cylinder $M = [0, 1/2] \times \mathbb{S}^1$ including into $N = [0, 1] \times \mathbb{S}^1$. – Eric Canton Dec 10 '19 at 2:19